Salt tank differential equation

[musicmar]

Homework Statement

Solve problem 23 under the assumption that the solution is pumped out at a faster rate of 10 gal/min. When is the tank empty?

23. A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.

The Attempt at a Solution

If I'm just looking for when the tank is empty, why do I need to care about salt? I solved it with a really simple D.E. using only the volume rates:

dV/dt = Rate in - Rate out

dV/dt = 5 gal/min - 10 gal/min ; V(0) = 500 gal

V = -5t + c

500 = 0 + c

c= 500

V = -5t + 500


0 = -5t + 500

t = 100 min.


I know this is the right answer for t, but the book also gives the equation for A(t) = 1000 - 10t - (1/10)(100-t)²

Is it really necessary to worry about the amount of salt, and if so, could someone help me?

Here is my work for solving for A(t):

dA/dt = 10 lb/min - (A(t)/500 gal) * (10 gal/min)

dA/dt = 5 - (1/50) A(t)

solve as linear; integrating factor = e^(1/50)t

(A*e^(1/50)t)' = 5*e^(1/50)t

A*e^(1/50)t = 250*e^(1/50)t + c
c = 500

A = 250 + 500*e^(1/50)t


Clearly, this is not the same answer. If someone could help me fix this, that would be great. Thank you.

 

[19SAI]

From the equation for Mixtures Problems:
dA/dt = input rate of salt - output rate of salt

From what I understand, when we base things on the first equation from problem 23: dA/dt = 10 – (A/100) , we obtain the DE A(t) = 1000 + ce^(−t/100).

But what seemed strange is that you assumed that A(0) is 500 gal, when it should be 0 gal, after all, the problem only said that 500 gal is the CAPACITY of the tank not the content of it at t=0. Therefore,

If A(0) = 0, then c = −1000 and A(t) = 1000−1000e^(−t/100).

On to your next problem, if its is pumping in by 5 gal/min while pumping out by 10 gal/min, we can assume that the new rate out is:
Rate out = 10A/500−(10−5) --(incorporate into the theoretical equation)-----> we obtain: dA/dt= 10−(10A/500−(10−5)) = 10− (2A/100− t)

so our new DE is : A(t) = 1000−10t +c(100− t)^2.
If A(0) = 0, then c = −1/10 .

The tank is empty in 100 minutes.

If this didn't help, I'm sorry. Yes the amount of salt is relevant is it is pumped into a tank filled with pure water and can therefore affect the rate of outflow of brine (salt+water) from the tank.

 

[Mark44]

From the equation for Mixtures Problems:
dA/dt = input rate of salt - output rate of salt

From what I understand, when we base things on the first equation from problem 23: dA/dt = 10 – (A/100) , we obtain the DE A(t) = 1000 + ce^(−t/100).

But what seemed strange is that you assumed that A(0) is 500 gal, when it should be 0 gal, after all, the problem only said that 500 gal is the CAPACITY of the tank not the content of it at t=0.

The problem statement says, "A large tank is filled to capacity with 500 gallons of pure water."

"Is filled" implies to me that the tank starts out with 500 gal at t = 0.

Therefore,

If A(0) = 0, then c = −1000 and A(t) = 1000−1000e^(−t/100).

On to your next problem, if its is pumping in by 5 gal/min while pumping out by 10 gal/min, we can assume that the new rate out is:
Rate out = 10A/500−(10−5) --(incorporate into the theoretical equation)-----> we obtain: dA/dt= 10−(10A/500−(10−5)) = 10− (2A/100− t)

so our new DE is : A(t) = 1000−10t +c(100− t)^2.
If A(0) = 0, then c = −1/10 .

The tank is empty in 100 minutes.

If this didn't help, I'm sorry.Yes the amount of salt is relevant is it is pumped into a tank filled with pure water and can therefore affect the rate of outflow of brine (salt+water) from the tank.

 

[BvU]

Dear mar,

your

dA/dt = 10 lb/min - (A(t)/500 gal) * (10 gal/min)

is based on a holdup of 500 gal at all times. But that isn't what's happening. The CST differential equation is different.